An elegant proof of Euler's identity! (1 Viewer)

[GeorgesBorges]

Don't ruin it for others and give everyone a go! PM me your answers =)
Any 4U student who knows their work should be able to do this question.... "should" being the key word

 

[Iruka]

Well, it is a nice question, but you have to assume that you can do calculus with complex numbers in the same way that you do calculus with real numbers.

 

[lolokay]

how about:

let y = cosx + isinx

a)show that dy/dx = iy
b)hence show that y = e^ix

does this make the same sort of assumption?

 

[Affinity]

same problem.. and you don't even have a definition for e^z where z is complex

 

[MC Squidge]

its not in the syllabus!

 

[GeorgesBorges]

Yup, calculus works in the complex field as well =)
treat "i" as just any algebraic constant.

 

[addikaye03]

b) You guys are just doing this right:

e^ix=cosx+isinx for any real number x

e^i(pi)=cos(pi)+isin(pi)

Since cos (pi)=-1, sin(pi)=0

e^i(pi)=-1

e^i(pi)+1=0

 

[Iruka]

Yup, calculus works in the complex field as well =)
treat "i" as just any algebraic constant.

Of course you can do it.

You are saying that good 4u students should be able to do it. My point is that 4u students have no justification for assuming that they can do it.

 

[Affinity]

Yup, calculus works in the complex field as well =)
treat "i" as just any algebraic constant.

There's a fundamental flaw to this question.

Remember you don't have a definition for e^z where z is complex. Which means that you will either need an explicit defintion for e^z to define the derivative or you can define exp(z) to be a function such that d/dz exp(z) = exp(z)

The first approach will probably mean defining e^z in terms of a power series, but then one can just check euler's identity by adding the power series.

the second approach will require you to prove that such a function in fact exists and that it agrees with what we know to be e^z when z is real.

Other approaches will require more machinery.


just to illustrate, the function f(x) = exp(-1/x) for x > 0 and 0 when x <= 0 is smooth (infinitely differentiable) when considered over the real numbers. but if you consider it as a complex number function then it has an essential discontinuity at 0

 

[dvse]

There's a fundamental flaw to this question.

Remember you don't have a definition for e^z where z is complex. Which means that you will either need an explicit defintion for e^z to define the derivative or you can define exp(z) to be a function such that d/dz exp(z) = exp(z)

I think we had a discussion on this somewhere before - the question clearly demonstrates that HSC maths is essentially harmful as the only thing it teaches people is to mindlessly apply rewrite rules to formal expressions without any understanding whatsoever.

 

[shaon0]

how about:

let y = cosx + isinx

a)show that dy/dx = iy
b)hence show that y = e^ix

does this make the same sort of assumption?

Yeah. That's how i would've done it.

 

[Joel8945]

Don't ruin it for others and give everyone a go! PM me your answers =)
Any 4U student who knows their work should be able to do this question.... "should" being the key word

a)


f'(x) = [((-sin(x)+icos(x)) x e^-ix) + ((-ie^-ix)x(cos(x)+sin(x))]

f'(x) = e^-ix[-sin(x)+icos(x)+sin(x)-icos(x)]

f'(x) = e^-ix x 0

f'(x) = 0

therefore f(x) = c, where c is a constant

but if |cos(x) + isin(x)| = 1 and -π ≤ x ≤ π => |e^-ix| = 1

=> f(x) = c = 1

therefore cos(x) + isin(x) = 1/(e^-ix) = e^ix

QED

b)

e^iπ + 1 = 0

e^iπ = cos(π) + isin(π) = -1 + i*0 = -1

therefore

e^iπ + 1 = LHS

-1 + 1 = LHs

0 = LHS

QED

Fun

 

[Joel8945]

a)

but if |cos(x) + isin(x)| = 1 and -π ≤ x ≤ π => |e^-ix| = 1

ok I should have stated that my assumption is that -π ≤ x ≤ π, because you need that to prove the proof. I have used these and understand the theory so my intuition knew what to expect but really the flaw to this proof is not giving x a restriction of values, but, you don't need to as e^-ix always has a magnitude of 1, but, by arriving at that conclusion you would have to assume that the proof you are proving is true! Unless I haven't looked at this from a different perspective where you can achieve that f(x) = 1 no matter what!

 

[Iruka]

Well, you could try x=0, couldn't you?

But the deeper problems still remain.